## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 5: Applications Of the Derivative Part 1 – Section 5.5: The Second-Derivative Test

5.5
The Second-Derivative Test

 1. The Second-Derivative Test

In Section 5.3 Theorem 4.1 we had the first-derivative test for local extrema. In this section we're going to study the
second-derivative test, which is also a test for local extrema.

Let f be a function that is twice differentiable near x1 and also near x2. See Fig. 1.1. Suppose f '(x1) = 0 and f ''(x1) > 0.
So the graph of f has a horizontal tangent at x1 and is concave up there. This means that f(x1) is a local minimum of f.
Similarly, if f '(x2) = 0 and f ''(x2) < 0, then f(x2) is a local maximum of f. # f(x1) is a local minimum and f(x2) is a local maximum of f.

Theorem 1.1    The Second-Derivative Test

 i.   If f '(x1) = 0 and f ''(x1) > 0, then f has a local minimum at x1. ii.  If f '(x1) = 0 and f ''(x1) < 0, then f has a local maximum at x1.

Proof

We prove part i. The proof of part ii is similar. Recalling that f '' is the derivative of f ' and using the definition of the
derivative we have: see note on the proof, below. Consider such small h's. If h < 0 then f '(x1 + h) < 0, and if h > 0 then f '(x1 + h) > 0.
Thus, by the first-derivative test, f has a local minimum at x1.

### Note On The Proof Because |h – 0| = |h| and f ''(x1)/2 > 0, the justification is complete.

 2. A Closer Look At The Second-Derivative Test

i.  The second-derivative test is a test for local extrema, not for inflection points. It has the same     purpose as the
first-derivative test.

ii.  The second-derivative test excludes points x1 where f '(x1) doesn't exist. As f '(x1) doesn't exist, f ''(x1) doesn't exist
either (if a function ( f ' in this case) isn't defined at x1, then it can't be differentiable there). So it's meaningless to talk
about the second derivative of f at x1.

iii.  The second-derivative test asserts nothing about what happens at points x1 where f '(x1) = 0 and f ''(x1) = 0 too.

Let's take a look at some examples. The graph of f1(x) = x4 is sketched in Fig. 2.1. We have f1'(0) = f1'(x)|x=0 =
4x3|x=0 = 0 and f1''(0) = f1''(x)|x=0 = 12x2|x=0 = 0. Clearly f1 has a local minimum at x = 0. The graph of f2(x) = –x4 is
sketched in Fig. 2.2. We have f2'(0) = f2'(x)|x=0 = – 4x3|x=0 = 0 and f2''(0) = f2''(x)|x=0 = –12x2|x=0 = 0. Clearly f2 has a
local maximum at x = 0. The graph of f3(x) = x3 is sketched in Fig. 2.3. We have f3'(0) = f3'(x)|x=0 = 3x2|x=0 = 0 and
f3''(0) = f3''(x)|x=0 = 6x|x=0 = 0. Clearly f3 has neither a local maximum nor a local minimum at x = 0, but it has an
inflection point there. Thus, if f '(x1) = 0 and f ''(x1) = 0, then f can have either a local maximum or a local minimum
or an inflection point at x1. Fig. 2.1   f1(x) = x4; f1'(0) = 0 and f1''(0) = 0; f1 has a local minimum at x = 0. Fig. 2.2   f2(x) = – x4; f2'(0) = 0 and f2''(0) = 0; f2 has a local maximum at x = 0. Fig. 2.3   f3(x) = x3; f3'(0) = 0 and f3''(0) = 0; f3 has an inflection point at x = 0.

 3. Using The Second-Derivative Test

Example 3.1

Use the second-derivative test to find and classify the critical points of f(x) = x – (1/2)x2xex. (The derivative of ex with
respect to x is ex itself.)

Solution

f '(x) = 1 – xex + xex = 1 – xex(1 – x) = (1 – x)(1 – ex);
f '(x) is defined everywhere;
f '(x) = 0 iff (1 – x)(1 – ex) = 0 iff (1 – x) = 0 or (1 – ex) = 0, so f '(x) = 0 at x = 1 and x = 0.

f ''(x) = – 1 + ex + exxex = – 1 + ex(2 – x);
f ''(1) = – 1 + e–1(2 – 1) = – 1 + 1/e < – 1 + 1 = 0;
f ''(0) = – 1 + 1(2 – 0) = 1 > 0.

So f has a local maximum at x = 1 and a local minimum at x = 0.

EOS

 4. Comparing The Second- To The First-Derivative Tests

i.  The second-derivative test is simpler than the first-derivative test if the second derivative isn't too complicated.

ii.  However, the second derivative may be so much more complicated than the first derivative that it may be less difficult
to use the first-derivative test than the second-derivative test.

iii.  The second-derivative test can't locate local extreme values at singular points ( points where the derivative doesn't
exist), while the first-derivative test can.

 Problems & Solutions

1.  Let f(x) = x/(1 + x2). Find and classify all local extrema of f using the second-derivative test.

Solution So f(1) = 1/2 is a local maximum and f(–1) = –1/2 is a local minimum of f. 2.  Let g(x) = x2ex. Find and classify all local extrema of g using the second-derivative test. (The derivative of ex with
respect to x is ex itself.)

#### Solution

g'(x) = 2xex + x2(–ex) = xex(2 – x);
g'(x) exists everywhere;

g'(x) = 0 at x = 0 and x = 2;
g''(x) = (ex + x(–ex))(2 – x) + xex(–1) = ex(2 – 4x + x2);
g''(0) = 2 > 0, g''(2) = –2e–2 < 0.

Thus by the second-derivative test g(0) = 0 is a local minimum and g(2) = 4e–2 is a local maximum of g. 3.  Let h(t) = t ln t. Find and classify all local extrema of h using the second-derivative test. (The derivative of ln x with
respect to x is 1/x.)

#### Solution

h'(t) = ln t + t(1/t) = ln t + 1;

h'(t) is defined for all t > 0;
h'(t) = 0 iff ln t = –1 iff t = e–1 = 1/e;
h''(t) = 1/t;
h''(1/e) = e > 0.
Consequently by the second-derivative test h(1/e) = (1/e) ln (1/e) = –1/e is a local minimum of h. 4.  The graph of the derivative f ' of a function f is shown below. Let G be the graph of f (not of f '). The second-derivative test is handy for some of the following questions.

a.  Does G has a local maximum at x = 0?
b.  Does G has a local maximum at x = –1?
c.  Does G has a local minimum at x = 1?
d.  Does G has a local minimum at x = 2?
e.  Does G has an inflection point at x = 0?
f.  Does G has an inflection point x = 1?
g.  Is G concave up on (0, 2)?
h.  Is G concave up on (1, 2)?

Solution

a.  Yes.
b.  No.
c.  No.
d.  Yes.
e.  No.
f.  Yes.
g.  No.
h.  Yes. 5.  Give an example of three functions f, g, and h such that f '(0) = f ''(0) = g'(0) = g''(0) = h'(0) = h''(0) = 0, but f has
a local minimum value, g has a local maximum value, and h has an inflection point at x = 0.

Solution

Let f(x) = x6, g(x) = –x6, and h(x) = x5. Note

For f and g, the second-derivative test doesn't apply but the first-derivative test does. For h, neither the first- nor the
second-derivative test applies; so we check to see if the given point is an inflection point.